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1
CHAPTER:
MATHEMATICAL INDUCTION
1
Introduction
2
Principle of Mathematical Induction
3
Problems Involving Summation
4
Problems Involving Differentiation
5
Problems Involving Sequences
6
Other Types of Problems
7
Miscellaneous Examples
 1 Introduction
Mathematical Induction is a method of proving a mathematical statement, theorem or formula.
n
n
For example, in summation, we have  r = (n + 1) where n is any positive integer.
2
r=1
We can use induction to prove this statement.
 2 Principle of Mathematical Induction
Let Pn be a statement involving a positive integer n. Suppose the following two conditions are satisfied:
(i) P1 is true (i.e. Pn is true for n = 1)
(ii) If Pk is true, then Pk + 1 is also true.
Then the statement Pn is true for all positive integers n.
There is a standard procedure for proving by Mathematical Induction with certain steps and statements.
Step 1 Let Pn be the statement, where n is any positive integer as stated in the question.
Step 2 Prove directly that P1 is correct.
If n does not start with 1 but some positive integer m, then prove that Pm is true.
Step 3 Assume Pk is true
Step 4 Prove Pk +1 is also true using step 3.
Step 5 Conclude that since Pk true implies Pk + 1 is also true, thus by mathematical induction, claim that Pn is
true for all positive integer values of n. If n does not start from 1 but some positive integer m, then use
induction to claim that Pn is true for all positive integer values of n  m.
Note: If the question does not state to prove by induction, you may use other methods to solve.
 3 Problems Involving Summation
To prove Pk + 1 is true, just add the (k + 1)th term to the Pk statement and simplify.
Example 3.1
n
Prove, by induction, that
r=1
Solution
1
 r2 = 12 + 22 + … + n2 = 6 n (n + 1)(2n + 1).
2
Example 3.2
n
Prove that
4
4r = 3 (4n – 1) by induction.
r=1
Solution
Example 3.3
Prove by induction that, for all values of the positive integer n, 2.1! + 5.2! + 10.3! +...+ (n2 + 1)n! = n(n+1)!.
Solution
3
Example 3.4
1
1
sin (n + 1)x sin nx
2
2
Prove that sinx + sin2x + sin3x + . . . + sin nx =
for n  1, n  +.
1
sin x
2
Solution
 4 Problems Involving Differentiation
To prove Pk + 1 is true, just differentiate the Pk statement one more time.
Example 4.1
Prove by induction that
Solution
d n
x = nx n  1 where n  +.
dx
4
Example 4.2
Prove by mathematical induction that given
y = xex , then
dny
= (x + n)ex.
dxn
Solution
Example 4.3 (AJC 99/1/17b)
Prove by induction that for all positive integers n,
Solution
dn
n
[sin (ax)] = an sin (ax +
) where a is a constant.
dxn
2
5
 5 Problems Involving Sequences
To prove that Pk + 1 is true, substitute the expression for Pk into Pk + 1 and simplify.
Example 5.1
If Un + 1 = 2Un + 2 for all positive integral values of n and U1 = 2, prove, by induction, that Un = 2(2n - 1).
Solution
Example 5.2 (TPJC 95/1/15a)
Given that a1 = 1 and an + 1 = 3an + 2n – 2 where n  1, n  N, prove by induction that an =
positive integers n.
Solution
3n
1
– n + for all
2
2
6
 6 Other Types of Problems
There is no standard way to solve this type of problems.
Example 6.1
Prove by induction that, when n  +, f(n) = 9n + 7 is exactly divisible by 8.
Solution
Example 6.2
Prove by induction that, when n  +, f(n) = 34n + 2 + 26n + 3 is exactly divisible by 17.
Solution
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 7 Miscellaneous Examples
Example 7.1 (YJC 01/1/4)
Prove by mathematical induction that, for all n  0,
n

(r2 + r + 1)(r!) = (n + 1)!(n + 1).
r=0
Solution
Example 7.2 (TJC 01/1/10)
Prove by induction that for all positive integers n,
integer.
Solution
dn
xe  ax = (1)n a n  1(ax  n)e ax where a is a positive
dxn
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SUMMARY (Mathematical Induction)
There is a standard procedure for proving by Mathematical Induction with certain steps and statements.
Step 1 Let Pn be the statement, where n is any positive integer as stated in the question.
Step 2 Prove directly that P1 is correct.
If n does not start with 1 but some positive integer m, then prove that P m is true.
Step 3 Assume Pk is true
Step 4 Prove Pk+1 is also true using step 3.
Step 5 Conclude that since Pk true implies Pk + 1 is also true, thus by mathematical induction, claim that P n is
true for all positive integer values of n. If n does not start from 1 but some positive integer m, then use
induction to claim that Pn is true for all positive integer values of n > m.
For problems involving summation, to prove P k + 1 is true, just add the (k + 1)th term to the Pk statement and
simplify.
For problems involving differentiation, to prove P k + 1 is true, just differentiate the Pk statement one more time.
For problems involving sequences, to prove that P k + 1 is true, substitute the expression for Pk into Pk + 1 and
simplify.
For other types of problems, there is no standard way to solve this type of problems.
Fun Facts!
111,111,111  111,111,111 = 12,345,678,987,654,321
"If I had eight hours to chop down a tree,
I'd spend six hours sharpening my axe."
Abraham Lincoln